Question: Is the function given below continuous/differentiable at $x=-2$ ? $f(x)=\begin{cases} -1.5x^2&,&x\leq -2 \\\\ 6x-5&,&x>-2 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Checking for continuity at $x=-2$ For the function to be continuous at $x=-2$, we need the two-sided limit $\lim_{x\to -2}f(x)$ to exist and be equal to $f(-2)$. This is the same as requiring that the two one-sided limits $\lim_{x\to -2^-}f(x)$ and $\lim_{x\to -2^+}f(x)$ exist and are equal to $f(-2)$. According to $f$ 's definition, $f(-2)=-1.5(-2)^2=-6$. $\lim_{x\to -2^-}f(x)$ $-1.5x^2$ evaluated at $x=-2$ is equal to $-6$. Since $-1.5x^2$ is continuous, we can be certain that $\lim_{x\to -2^-}f(x)=-6$. $\lim_{x\to -2^+}f(x)$ $6x-5$ evaluated at $x=-2$ is equal to $-17$. Since $6x-5$ is continuous, we can be certain that $\lim_{x\to -2^+}f(x)=-17$. The two limits exits, but they are not equal. Therefore, the function is not continuous at $x=-2$. Graphically, the function skips a step at this point. [I would like to see that, please!] Checking for differentiability at $x=-2$ Since the function isn't continuous at $x=-2$, it cannot be differentiable at that point. In conclusion, the function is neither continuous nor differentiable at $x=-2$.